Question

$\int {\tan }^{3}2x\sec 2xdx$

Answer 1

$LetI=\int {\tan }^{3}2x.\sec 2xdx$

$=\int {\tan }^{2}2x.\sec 2x.\tan 2xdx$

$=\int ({\sec }^{2}2x-1)\sec 2x.\tan 2xdx$

Put $\sec 2x=t$

$2\sec 2x.\tan 2xdx=dt$

Now $\sec 2x.\tan 2xdx=\frac{dt}{2}$

$I=\frac{1}{2}\int (t^2-1)dt$

$=\frac{1}{2}[\frac{t^3}{3}-t]+C$

$=\frac{1}{2}[\frac{{\sec }^{3}2x}{3}-\sec 2x]+C$